What is the extraneous solution to these equations? $\dfrac{x^2 + 41}{x - 10} = \dfrac{18x - 39}{x - 10}$
Answer: Multiply both sides by $x - 10$ $ \dfrac{x^2 + 41}{x - 10} (x - 10) = \dfrac{18x - 39}{x - 10} (x - 10)$ $ x^2 + 41 = 18x - 39$ Subtract $18x - 39$ from both sides: $ x^2 + 41 - (18x - 39) = 18x - 39 - (18x - 39)$ $ x^2 + 41 - 18x + 39 = 0$ $ x^2 + 80 - 18x = 0$ Factor the expression: $ (x - 10)(x - 8) = 0$ Therefore $x = 10$ or $x = 8$ At $x = 10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 10$, it is an extraneous solution.